Interpretation of basis of tangent space

Question

Suppose a regular surface $S\subset \Bbb R^3$ parametrized by $\Sigma:U\to \Bbb R^3:(x_1,x_2)\mapsto (x(x_1,x_2),y(x_1,x_2),z(x_1,x_2))$. If $p=\Sigma(u_0,v_0)\in S$, then the tangent space to $S$ at $p$ is

$$T_pS=\{\lambda \frac {\partial \Sigma}{\partial x_1}(u_0,v_0)+\mu\frac {\partial \Sigma}{\partial x_2}(u_0,v_0)\mid \lambda,\mu\in \Bbb R \},$$

and we can rewrite equivalently

$$T_pS=\{(\lambda \frac {\partial}{\partial x_1}|_p+\mu\frac {\partial}{\partial x_2}|_p)\Sigma\mid \lambda,\mu\in \Bbb R\}.$$

Τhe question is how this expression agrees exactly with the fact that $T_pM^2={\rm span}\{\frac {\partial}{\partial x_1}|_p, \frac {\partial}{\partial x_2}|_p\}$, where $M^2$ is a 2-d differential manifold, provided that in our example $S=M^2$? (More precisely: The basis of $T_pS$ for classical approach is a span of operators acting on function, but for riemannian approach is a span of these operators alone.)

The whole question is about how could we interpret the basis $\{\frac {\partial}{\partial x_i}|_p\}_i$ in a "common" way.

Answer 1

Your equation $T_pM^2=$span$\{\frac {\partial}{\partial x_1}, \frac {\partial}{\partial x_2}\}$ is not quite right because the the point $p \in \Sigma(U)$ is a parameter for the operators $\frac{\partial}{\partial x_i}$. Once you correctly apply those operators at the particular point $p=\Sigma(u_0,v_0)$ then you'll get what you want, namely $$T_p M^2= \text{span} \bigl\{\frac {\partial}{\partial x_1} \mid_p , \frac {\partial}{\partial x_2} \mid_p \bigr\} = \left\{\lambda \frac {\partial}{\partial x_1} \mid_p + \mu \frac {\partial}{\partial x_2} \mid_p : \lambda,\mu \in \mathbb R \right\} $$

Answer 2

I think that I found the key of this question: the view of a vector as a function applied on functions.

Let $a\in \Bbb R^n$, $v_p\in T_p\Bbb R^n$ and $f: \Bbb R^n \to \Bbb R$ a $C^1$ -at least- function. We are familiar with the directional derivative of $f$

$$v_p(f):=\frac{d}{dt}f(a+tv)|_{t=0}=\nabla f(a+tv)\cdot v|_{t=0}.$$

For $v_p:=(e_i)_p$ we have

$$(e_i)_p(f)=\nabla f(a+te_i)\cdot e_i|_{t=0}=\frac{\partial f}{\partial x_i}(p):=(\frac {\partial}{\partial x_i}|_p)(f).$$

The function $f$ is arbitrary, so we can identify

$$(e_i)_p\equiv \frac {\partial}{\partial x_i}|_p.$$

This means that we can think of $\frac {\partial}{\partial x_i}|_p$ as the unit basis vector of $T_p\Bbb R^n$.